What is the integral of x^2(x^3+7)^6 ?

The integral of x^2(x^3+7)^6 is 1/21 (x^3+7)^7+C

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integral of x^2(x^3+7)^6

\int x^{2} (x^{3} + 7)^{6} d x

\frac{1}{21} (x^{3} + 7)^{7} + C

Solution steps

\int x^{2} (x^{3} + 7)^{6} d x

Apply u-substitution

: \int \frac{u ^{6}}{3} d u

= \int \frac{u ^{6}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{1}{3} \cdot \frac{u ^{7}}{7}

Substitute back

u = x^{3} + 7

= \frac{1}{3} \cdot \frac{( x ^{3} + 7 ) ^{7}}{7}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} + 7 ) ^{7}}{7} : \frac{1}{21} (x^{3} + 7)^{7}

= \frac{1}{21} (x^{3} + 7)^{7}

Add a constant to the solution

= \frac{1}{21} (x^{3} + 7)^{7} + C

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What is the integral of x^2(x^3+7)^6 ?
The integral of x^2(x^3+7)^6 is 1/21 (x^3+7)^7+C